Momentum + coefficient of restitution this is driving me nuts

[ar202]

Momentum + coefficient of restitution this is driving me nuts!

Homework Statement

A Mass of 2kg traveling at 2m/s is struck by a mass of 4 kg traveling at 3m/s. If the coefficient of restitution for the two materials is 0.4 calculate their final velocities.

Homework Equations

M₁U₁+M₂U₂=M₁V₁+M₂V₂

E=V₂-V₁/U₂-U₁

The Attempt at a Solution

From what i know, the best way to do it is use both equations and create two simultaneous equations?

Is this correct? i seem to end up with V₂=-1.2 however I am pretty sure this is wrong.

any help would be great.

 

[tiny-tim]

Welcome to PF!

A Mass of 2kg traveling at 2m/s is struck by a mass of 4 kg traveling at 3m/s. If the coefficient of restitution for the two materials is 0.4 calculate their final velocities.
…
i seem to end up with V₂=-1.2 however I am pretty sure this is wrong.

Hi ar202! Welcome to PF!

Show us your full calculations, and then we'll see what went wrong, and we'll know how to help!

 

[ar202]

Cheers dude :-)

first bit...

2*2+4*(-4)=2*V1+4*V2

-12=2*V1+4*V2

second bit...

0.4*2-0.4*4=v2-v1

2.4=v2-v1

so then i have

-12=2*V1+4*V2
+
4.8=2V2-2V1

which gives

-7.2=6V₂...

 

[tiny-tim]

Hi ar202!

erm … you keep writing 4 instead of 3 …

A Mass of 2kg traveling at 2m/s is struck by a mass of 4 kg traveling at 3m/s. If the coefficient of restitution for the two materials is 0.4 calculate their final velocities.

2*2+4*(-4)=2*V1+4*V2

…

0.4*2-0.4*4=v2-v1

 

[ar202]

Hi ar202!

erm … you keep writing 4 instead of 3 …

haha so i have :p

the answer is still wrong... any tips?

 

[tiny-tim]

Have you checked that your signs for V1 and V2 are the right way round in both equations?

(and btw, were they traveling in the same direction originally, or in opposite directions?)

 

[ar202]

Have you checked that your signs for V1 and V2 are the right way round in both equations?

(and btw, were they traveling in the same direction originally, or in opposite directions?)

it doesn't mention directions, just that it is struck. I've found the answers - V1=2.53 & V2=2.93

still struggling to get to those answers.

 

[tiny-tim]

it doesn't mention directions, just that it is struck. I've found the answers - V1=2.53 & V2=2.93

still struggling to get to those answers.

Well, then they're obviously going in the same direction originally …

relative speed before = 1, after = 0.4

 

[ar202]

Well, then they're obviously going in the same direction originally …

relative speed before = 1, after = 0.4

Ok so using the same method as before i can get V2=2.83 & V1=2.34

am i going about it the right way...?

 

[tiny-tim]

hmm … relative speed after = 0.49 … not close enough to 0.4

you must be making a mistake somewhere, but without seeing the details, I can't say where.

 

[ar202]

hmm … relative speed after = 0.49 … not close enough to 0.4

you must be making a mistake somewhere, but without seeing the details, I can't say where.

first part

2*2+4*3=2*v1+4*v2

0.4(3-2)=v2-v1 X 2 then I've added it to the first part...

so 18=6*v2

18/6=3 ... actually my answer was 3 for v2. I must be more tired than i thought :p

 

[tiny-tim]

oooh, i see it now …

0.4(3-2)=v2-v1 X 2

… there's no "X 2" …

the coefficient of restitution is just |u1 - u2|/|v1 - v2|

 

[ar202]

the 'X2' was the simultaneous equation part...

 

[tiny-tim]

the 'X2' was the simultaneous equation part...

(just got up :zzz: …)

oh, it multiplied the whole equation!

ok, then your 18 should be 16.8 ​

 

[ar202]

(just got up :zzz: …)

oh, it multiplied the whole equation!

ok, then your 18 should be 16.8 ​

ok i finally got there

got it down to 6v2=15.8

15.8/6=2.53333333333

thanks for your help dude. think i made it a lot harder than it should have been!

 

[tiny-tim]

15.8/6=2.53333333333

you do know that 15.2/6=2.53333333333 ?